Higher November 2020 Paper 1 Q21
21 Express
\(\dfrac{1}{3x - 2} \times \dfrac{9x^2 - 4}{3x^2 - 13x - 10} - \dfrac{7}{x - 1}\)
as a single fraction in its simplest form.
(5)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{9x^2 - 4}{3x^2 - 13x - 10} =\right) \dfrac{(3x + 2)(3x - 2)}{(3x + 2)(x - 5)}\) | M1 |
| \(\left(\dfrac{9x^2 - 4}{3x^2 - 13x - 10} =\right) \dfrac{(3x + 2)(3x - 2)}{(3x + 2)(x - 5)}\) | M1 |
| E.g. of denominators \((3x - 2)(3x^2 - 13x - 10)(x - 1)\) or \((3x - 2)(3x + 2)(x - 5)(x - 1)\) or \(9x^4 - 54x^3 + 41x^2 + 24x - 20\) or \((3x + 2)(x - 5)(x - 1)\) or \(3x^3 - 16x^2 + 3x + 10\) or \((3x - 2)(x - 5)(x - 1)\) or \(3x^3 - 20x^2 + 27x - 10\) or \((x - 5)(x - 1)\) or \(x^2 - 6x + 5\) | M1 |
\(\dfrac{x - 1 - 7(x - 5)}{(x - 5)(x - 1)}\) or \(\dfrac{x - 1 - 7x + 35}{(x - 5)(x - 1)}\) or \(\dfrac{x - 1 - 7(x - 5)}{x^2 - 6x + 5}\) or \(\dfrac{x - 1 - 7x + 35}{x^2 - 6x + 5}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{2(17 - 3x)}{(x - 5)(x - 1)}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for either
\((3x + 2)(3x - 2)\) or \((3x + 2)(x - 5)\)
M2 for \(\dfrac{9x^2 - 4}{(9x^2 - 4)(x - 5)} = \dfrac{1}{(x - 5)}\)
M1: for
\((3x + 2)(3x - 2)\)
and
\((3x + 2)(x - 5)\)
M1: (indep) ft their fractions for use of a correct common denominator for 2 fractions with algebraic denominators
NB: fractions need not be simplified
M1: for a correct fraction with a correct quadratic denominator – may or may not be expanded which leads to a correct answer
A1: accept \(\dfrac{34 - 6x}{(x - 5)(x - 1)}\) oe; if denominator is expanded then it must be correct